# Maximum length of same indexed subarrays from two given arrays satisfying the given condition

• Difficulty Level : Expert
• Last Updated : 11 Jul, 2022

Given two arrays arr[] and brr[] and an integer C, the task is to find the maximum possible length, say K, of the same indexed subarrays such that the sum of the maximum element in the K-length subarray in brr[] with the product between K and sum of the K-length subarray in arr[] does not exceed C.

Examples:

Input: arr[] = {2, 1, 3, 4, 5}, brr[] = {3, 6, 1, 3, 4}, C = 25
Output: 3
Explanation: Considering the subarrays arr[] = {2, 1, 3} (Sum = 6) and brr[] = {3, 6, 1} (Maximum element = 6), Maximum element + sum * K = 6 + 6 * 3 = 24, which is less than C(= 25).

Input: arr[] ={1, 2, 1, 6, 5, 5, 6, 1}, brr[] = {14, 8, 15, 15, 9, 10, 7, 12}, C = 40
Output: 3
Explanation: Considering the subarrays arr[] = {1, 2, 1} (Sum = 4) and brr[] = {14, 8, 15} (Maximum element = 6), Maximum element + sum * K = 15 + 4 * 3 = 27, which is less than C(= 40).

Naive Approach: The simplest approach is to generate all possible subarrays of the two given arrays and consider all similarly indexed subarrays from both the arrays and check for the given condition. Print the maximum length of subarrays satisfying the given conditions.

Time Complexity: O(K*N2)
Auxiliary Space: O(1)

Binary-Search-based Approach: To optimize the above approach, the idea is to use Binary Search to find the possible value of K and to find the sum of each subarray of length K using the Sliding Window Technique. Follow the steps below to solve the problem:

• Build a Segment Tree to find the maximum value among all possible ranges.
• Perform Binary Search over the range [0, N] to find the maximum possible size of the subarray.
• Initialize low as 0 and high as N.
• Find the value of mid as (low + high)/2.
• Check if it is possible to get the maximum size of the subarray as mid or not by checking the given condition. If found to be true, then update the maximum length as mid and low as (mid + 1).
• Otherwise, update high as (mid – 1).
• After completing the above steps, print the value of the maximum length stored.

Below is the implementation of the above approach:

## C++14

 // C++ program for the above approach #include  using namespace std;   // Stores the segment tree node valuesint seg;  // Function to find maximum element// in the given rangeint getMax(int b[], int ss, int se, int qs,           int qe, int index){          // If the query is out of bounds    if (se < qs || ss > qe)        return INT_MIN / 2;      // If the segment is completely    // inside the query range    if (ss >= qs && se <= qe)        return seg[index];      // Calculate the mid    int mid = ss + (se - ss) / 2;      // Return maximum in left & right    // of the segment tree recursively    return max(        getMax(b, ss, mid, qs,               qe, 2 * index + 1),        getMax(b, mid + 1, se,               qs, qe, 2 * index + 2));}  // Function to check if it is possible// to have such a subarray of length Kbool possible(int a[], int b[], int n,              int c, int k){    int sum = 0;          // Check for first window of size K    for(int i = 0; i < k; i++)     {        sum += a[i];    }       // Calculate the total cost and    // check if less than equal to c    int total_cost = sum * k + getMax(                     b, 0, n - 1,                         0, k - 1, 0);       // If it satisfy the condition    if (total_cost <= c)        return true;       // Find the sum of current subarray    // and calculate total cost    for(int i = k; i < n; i++)    {                  // Include the new element        // of current subarray        sum += a[i];           // Discard the element        // of last subarray        sum -= a[i - k];           // Calculate total cost        // and check <=c        total_cost = sum * k + getMax(                     b, 0, n - 1,                       i - k + 1, i, 0);           // If possible, then        // return true        if (total_cost <= c)            return true;    }       // If it is not possible    return false;}  // Function to find maximum length// of subarray such that sum of// maximum element in subarray in brr[] and// sum of subarray in arr[] * K is at most Cint maxLength(int a[], int b[], int n, int c){          // Base Case    if (n == 0)        return 0;      // Let maximum length be 0    int max_length = 0;       int low = 0, high = n;       // Perform Binary search    while (low <= high)    {                  // Find mid value        int mid = low + (high - low) / 2;                  // Check if the current mid        // satisfy the given condition        if (possible(a, b, n, c, mid) != false)        {                          // If yes, then store length            max_length = mid;            low = mid + 1;        }           // Otherwise        else            high = mid - 1;    }       // Return maximum length stored    return max_length;}   // Function that builds segment Treevoid build(int b[], int index, int s, int e){          // If there is only one element    if (s == e)     {        seg[index] = b[s];        return;    }       // Find the value of mid    int mid = s + (e - s) / 2;      // Build left and right parts    // of segment tree recursively    build(b, 2 * index + 1, s, mid);    build(b, 2 * index + 2, mid + 1, e);      // Update the value at current    // index    seg[index] = max(        seg[2 * index + 1],        seg[2 * index + 2]);}  // Function that initializes the// segment Treevoid initialiseSegmentTree(int N){    int seg[4 * N];}    // Driver Code int main() {     int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };    int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };           int C = 40;           int N = sizeof(A) / sizeof(A);           // Initialize and Build the    // Segment Tree    initialiseSegmentTree(N);    build(B, 0, 0, N - 1);           // Function Call    cout << (maxLength(A, B, N, C));}   // This code is contributed by susmitakundugoaldanga

## Java

 // Java program for the above approach  import java.io.*;import java.util.*;class GFG {      // Stores the segment tree node values    static int seg[];      // Function to find maximum length    // of subarray such that sum of    // maximum element in subarray in brr[] and    // sum of subarray in arr[] * K is at most C    public static int maxLength(        int a[], int b[], int n, int c)    {        // Base Case        if (n == 0)            return 0;          // Let maximum length be 0        int max_length = 0;          int low = 0, high = n;          // Perform Binary search        while (low <= high) {              // Find mid value            int mid = low + (high - low) / 2;              // Check if the current mid            // satisfy the given condition            if (possible(a, b, n, c, mid)) {                  // If yes, then store length                max_length = mid;                low = mid + 1;            }              // Otherwise            else                high = mid - 1;        }          // Return maximum length stored        return max_length;    }      // Function to check if it is possible    // to have such a subarray of length K    public static boolean possible(        int a[], int b[], int n,        int c, int k)    {        int sum = 0;          // Check for first window of size K        for (int i = 0; i < k; i++) {            sum += a[i];        }          // Calculate the total cost and        // check if less than equal to c        int total_cost            = sum * k + getMax(b, 0, n - 1,                               0, k - 1, 0);          // If it satisfy the condition        if (total_cost <= c)            return true;          // Find the sum of current subarray        // and calculate total cost        for (int i = k; i < n; i++) {              // Include the new element            // of current subarray            sum += a[i];              // Discard the element            // of last subarray            sum -= a[i - k];              // Calculate total cost            // and check <=c            total_cost                = sum * k                  + getMax(b, 0, n - 1,                           i - k + 1, i, 0);              // If possible, then            // return true            if (total_cost <= c)                return true;        }          // If it is not possible        return false;    }      // Function that builds segment Tree    public static void build(        int b[], int index, int s, int e)    {        // If there is only one element        if (s == e) {            seg[index] = b[s];            return;        }          // Find the value of mid        int mid = s + (e - s) / 2;          // Build left and right parts        // of segment tree recursively        build(b, 2 * index + 1, s, mid);        build(b, 2 * index + 2, mid + 1, e);          // Update the value at current        // index        seg[index] = Math.max(            seg[2 * index + 1],            seg[2 * index + 2]);    }      // Function to find maximum element    // in the given range    public static int getMax(        int b[], int ss, int se, int qs,        int qe, int index)    {        // If the query is out of bounds        if (se < qs || ss > qe)            return Integer.MIN_VALUE / 2;          // If the segment is completely        // inside the query range        if (ss >= qs && se <= qe)            return seg[index];          // Calculate the mid        int mid = ss + (se - ss) / 2;          // Return maximum in left & right        // of the segment tree recursively        return Math.max(            getMax(b, ss, mid, qs,                   qe, 2 * index + 1),            getMax(b, mid + 1, se,                   qs, qe, 2 * index + 2));    }      // Function that initializes the    // segment Tree    public static void    initialiseSegmentTree(int N)    {        seg = new int[4 * N];    }      // Driver Code    public static void main(String[] args)    {        int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };        int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };          int C = 40;          int N = A.length;          // Initialize and Build the        // Segment Tree        initialiseSegmentTree(N);        build(B, 0, 0, N - 1);          // Function Call        System.out.println(maxLength(A, B, N, C));    }}

## Python3

 # Python3 program for the above approach import math   # Stores the segment tree node valuesseg = [0 for x in range(10000)] INT_MIN = int(-10000000)  # Function to find maximum element# in the given rangedef getMax(b, ss, se, qs, qe, index):          # If the query is out of bounds    if (se < qs or ss > qe):        return int(INT_MIN / 2)              # If the segment is completely    # inside the query range    if (ss >= qs and se <= qe):        return seg[index]              # Calculate the mid    mid = int(int(ss) + int((se - ss) / 2))          # Return maximum in left & right    # of the segment tree recursively    return max(getMax(b, ss, mid, qs,                      qe, 2 * index + 1),               getMax(b, mid + 1, se, qs,                      qe, 2 * index + 2))  # Function to check if it is possible# to have such a subarray of length Kdef possible(a,  b, n, c, k):          sum = int(0)          # Check for first window of size K    for i in range(0, k):        sum += a[i]              # Calculate the total cost and    # check if less than equal to c    total_cost = int(sum * k +               getMax(b, 0, n - 1,                        0, k - 1, 0))       # If it satisfy the condition    if (total_cost <= c):        return 1      # Find the sum of current subarray    # and calculate total cost    for i in range (k, n):                  # Include the new element        # of current subarray        sum += a[i]                  # Discard the element        # of last subarray        sum -= a[i - k]          # Calculate total cost        # and check <=c        total_cost = int(sum * k + getMax(               b, 0, n - 1,i - k + 1, i, 0))           # If possible, then        # return true        if (total_cost <= c):            return 1                  # If it is not possible    return 0  # Function to find maximum length# of subarray such that sum of# maximum element in subarray in brr[] and# sum of subarray in arr[] * K is at most Cdef maxLength(a, b, n, c):          # Base Case    if (n == 0):        return 0      # Let maximum length be 0    max_length = int(0)       low = 0    high = n          # Perform Binary search    while (low <= high):                  # Find mid value        mid = int(low + int((high - low) / 2))                  # Check if the current mid        # satisfy the given condition        if (possible(a, b, n, c, mid) != 0):                          # If yes, then store length            max_length = mid            low = mid + 1                      # Otherwise        else:            high = mid - 1       # Return maximum length stored    return max_length   # Function that builds segment Treedef build(b, index, s, e):          # If there is only one element    if (s == e):        seg[index] = b[s]        return          # Find the value of mid    mid = int(s + int((e - s) / 2))      # Build left and right parts    # of segment tree recursively    build(b, 2 * index + 1, s, mid)    build(b, 2 * index + 2, mid + 1, e)      # Update the value at current    # index    seg[index] = max(seg[2 * index + 1],                     seg[2 * index + 2])  #  Driver Code A = [ 1, 2, 1, 6, 5, 5, 6, 1 ]B = [ 14, 8, 15, 15, 9, 10, 7, 12 ]       C = int(40)N = len(A)    # Initialize and Build the# Segment Treebuild(B, 0, 0, N - 1)  # Function Callprint((maxLength(A, B, N, C)))  # This code is contributed by Stream_Cipher

## C#

 // C# program for the above approach using System;  class GFG{      // Stores the segment tree node values static int[] seg;   // Function to find maximum length // of subarray such that sum of // maximum element in subarray in brr[] and // sum of subarray in arr[] * K is at most C static int maxLength(int[] a, int[] b,                     int n, int c) {           // Base Case     if (n == 0)         return 0;       // Let maximum length be 0     int max_length = 0;       int low = 0, high = n;       // Perform Binary search     while (low <= high)    {                   // Find mid value         int mid = low + (high - low) / 2;           // Check if the current mid         // satisfy the given condition         if (possible(a, b, n, c, mid))        {                           // If yes, then store length             max_length = mid;             low = mid + 1;         }           // Otherwise         else            high = mid - 1;     }       // Return maximum length stored     return max_length; }   // Function to check if it is possible // to have such a subarray of length K static bool possible(int[] a, int[] b, int n,                     int c, int k) {     int sum = 0;       // Check for first window of size K     for(int i = 0; i < k; i++)    {        sum += a[i];     }       // Calculate the total cost and     // check if less than equal to c     int total_cost = sum * k +              getMax(b, 0, n - 1,                         0, k - 1, 0);       // If it satisfy the condition     if (total_cost <= c)         return true;       // Find the sum of current subarray     // and calculate total cost     for(int i = k; i < n; i++)    {                   // Include the new element         // of current subarray         sum += a[i];           // Discard the element         // of last subarray         sum -= a[i - k];           // Calculate total cost         // and check <=c         total_cost = sum * k +              getMax(b, 0, n - 1,                        i - k + 1, i, 0);           // If possible, then         // return true         if (total_cost <= c)             return true;     }       // If it is not possible     return false; }   // Function that builds segment Tree static void build(int[] b, int index,                  int s, int e) {           // If there is only one element     if (s == e)     {         seg[index] = b[s];         return;     }       // Find the value of mid     int mid = s + (e - s) / 2;       // Build left and right parts     // of segment tree recursively     build(b, 2 * index + 1, s, mid);     build(b, 2 * index + 2, mid + 1, e);       // Update the value at current     // index     seg[index] = Math.Max(         seg[2 * index + 1],         seg[2 * index + 2]); }   // Function to find maximum element // in the given range public static int getMax(int[] b, int ss,                          int se, int qs,                          int qe, int index) {           // If the query is out of bounds     if (se < qs || ss > qe)         return Int32.MinValue / 2;       // If the segment is completely     // inside the query range     if (ss >= qs && se <= qe)         return seg[index];       // Calculate the mid     int mid = ss + (se - ss) / 2;       // Return maximum in left & right     // of the segment tree recursively     return Math.Max(         getMax(b, ss, mid, qs,                qe, 2 * index + 1),         getMax(b, mid + 1, se,                qs, qe, 2 * index + 2)); }   // Function that initializes the // segment Tree static void initialiseSegmentTree(int N) {     seg = new int[4 * N]; }   // Driver Code    static void Main(){    int[] A = { 1, 2, 1, 6, 5, 5, 6, 1 };     int[] B = { 14, 8, 15, 15, 9, 10, 7, 12 };       int C = 40;       int N = A.Length;       // Initialize and Build the     // Segment Tree     initialiseSegmentTree(N);     build(B, 0, 0, N - 1);       // Function Call     Console.WriteLine(maxLength(A, B, N, C)); }}  // This code is contributed by divyesh072019

## Javascript

 

Output:

3

Time Complexity: O(N*(log N)2)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use a Deque by using a monotone queue such that for each subarray of fixed length we can find a maximum in O(1) time. For any subarray in the range [i, i + K – 1] the value expression to be calculated is given by: Below are the steps:

• Perform the Binary Search over the range [0, N] to find the maximum possible size of the subarray.
• Initialize low as 0 and high as N.
• Find the value of mid as (low + high)/2.
• Check if it is possible to get the maximum size of the subarray as mid or not as:
• Use deque to find the maximum element in each subarray of size K in the array brr[].
• Find the value of the expression and if it at most C then break out of this condition.
• Else check for all possible subarray size mid and if the value of the expression and if it at most C then break out of this condition.
• Return false if none of the above conditions satisfies.
• If the current mid satisfies the given conditions then update maximum length as mid and low as (mid + 1).
• Else Update high as (mid – 1).
• After the above steps, print the value of the maximum length stored.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std;  // Function to check if it is possible// to have such a subarray of length Kbool possible(int a[], int b[], int n, int c,                        int k){      // Finds the maximum element    // in each window of size k    deque <int> dq;      // Check for window of size K    int sum = 0;      // For all possible subarrays of    // length k    for (int i = 0; i < k; i++)     {        sum += a[i];          // Until deque is empty        while (dq.size() > 0 && b[i] > b[dq.back()])            dq.pop_back();        dq.push_back(i);    }      // Calculate the total cost and    // check if less than equal to c    int total_cost = sum * k + b[dq.front()];    if (total_cost <= c)        return true;      // Find sum of current subarray    // and the total cost    for (int i = k; i < n; i++)    {          // Include the new element        // of current subarray        sum += a[i];          // Discard the element        // of last subarray        sum -= a[i - k];          // Remove all the elements        // in the old window        while (dq.size() > 0 && dq.front() <= i - k)            dq.pop_front();        while (dq.size() > 0 && b[i] > b[dq.back()])            dq.pop_back();              dq.push_back(i);          // Calculate total cost        // and check <=c        total_cost = sum * k + b[dq.front()];          // If current subarray        // length satisfies        if (total_cost <= c)            return true;    }      // If it is not possible    return false;}  // Function to find maximum length// of subarray such that sum of// maximum element in subarray in brr[] and// sum of subarray in arr[] * K is at most Cint maxLength(int a[], int b[], int n, int c){        // Base Case    if (n == 0)        return 0;      // Let maximum length be 0    int max_length = 0;    int low = 0, high = n;      // Perform Binary search    while (low <= high)    {          // Find mid value        int mid = low + (high - low) / 2;          // Check if the current mid        // satisfy the given condition        if (possible(a, b, n, c, mid))         {              // If yes, then store length            max_length = mid;            low = mid + 1;        }          // Otherwise        else            high = mid - 1;    }      // Return maximum length stored    return max_length;}  // Driver Codeint main(){      int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };    int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };    int N = sizeof(A)/sizeof(A);    int C = 40;    cout << maxLength(A, B, N, C);    return 0;}  // This code is contributed by Dharanendra L V

## Java

 // Java program for the above approach  import java.io.*;import java.util.*;class GFG {      // Function to find maximum length    // of subarray such that sum of    // maximum element in subarray in brr[] and    // sum of subarray in arr[] * K is at most C    public static int maxLength(        int a[], int b[], int n, int c)    {        // Base Case        if (n == 0)            return 0;          // Let maximum length be 0        int max_length = 0;          int low = 0, high = n;          // Perform Binary search        while (low <= high) {              // Find mid value            int mid = low + (high - low) / 2;              // Check if the current mid            // satisfy the given condition            if (possible(a, b, n, c, mid)) {                  // If yes, then store length                max_length = mid;                low = mid + 1;            }              // Otherwise            else                high = mid - 1;        }          // Return maximum length stored        return max_length;    }      // Function to check if it is possible    // to have such a subarray of length K    public static boolean possible(        int a[], int b[], int n, int c, int k)    {          // Finds the maximum element        // in each window of size k        Deque dq            = new LinkedList();          // Check for window of size K        int sum = 0;          // For all possible subarrays of        // length k        for (int i = 0; i < k; i++) {              sum += a[i];              // Until deque is empty            while (dq.size() > 0                   && b[i] > b[dq.peekLast()])                dq.pollLast();            dq.addLast(i);        }          // Calculate the total cost and        // check if less than equal to c        int total_cost = sum * k                         + b[dq.peekFirst()];        if (total_cost <= c)            return true;          // Find sum of current subarray        // and the total cost        for (int i = k; i < n; i++) {              // Include the new element            // of current subarray            sum += a[i];              // Discard the element            // of last subarray            sum -= a[i - k];              // Remove all the elements            // in the old window            while (dq.size() > 0                   && dq.peekFirst()                          <= i - k)                dq.pollFirst();              while (dq.size() > 0                   && b[i]                          > b[dq.peekLast()])                dq.pollLast();              dq.add(i);              // Calculate total cost            // and check <=c            total_cost = sum * k                         + b[dq.peekFirst()];              // If current subarray            // length satisfies            if (total_cost <= c)                return true;        }          // If it is not possible        return false;    }      // Driver Code    public static void main(String[] args)    {        int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };        int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };          int N = A.length;          int C = 40;          System.out.println(            maxLength(A, B, N, C));    }}

## Python3

 # Python3 program for the above approach  # Function to find maximum length# of subarray such that sum of# maximum element in subarray in brr[] and# sum of subarray in []arr * K is at most Cdef maxLength(a, b, n, c):      # Base Case    if(n == 0):        return 0      # Let maximum length be 0    max_length = 0    low = 0    high = n      # Perform Binary search    while(low <= high):          # Find mid value        mid = int(low + (high - low) / 2)          # Check if the current mid        # satisfy the given condition        if(possible(a, b, n, c, mid)):              # If yes, then store length            max_length = mid            low = mid + 1          # Otherwise        else:            high = mid - 1      # Return maximum length stored    return max_length  # Function to check if it is possible# to have such a subarray of length Kdef possible(a, b, n, c, k):      # Finds the maximum element    # in each window of size k    dq = []          # Check for window of size K    Sum = 0      # For all possible subarrays of    # length k    for i in range(k):        Sum += a[i]          # Until deque is empty        while(len(dq) > 0 and b[i] > b[dq[len(dq) - 1]]):            dq.pop(len(dq) - 1)        dq.append(i)      # Calculate the total cost and    # check if less than equal to c    total_cost = Sum * k + b[dq]    if(total_cost <= c):        return True      # Find sum of current subarray    # and the total cost    for i in range(k, n):          # Include the new element        # of current subarray        Sum += a[i]          # Discard the element        # of last subarray        Sum -= a[i - k]          # Remove all the elements        # in the old window        while(len(dq) > 0 and dq <= i - k):            dq.pop(0)        while(len(dq) > 0 and b[i] > b[dq[len(dq) - 1]]):            dq.pop(len(dq) - 1)        dq.append(i)          # Calculate total cost        # and check <=c        total_cost = Sum * k + b[dq]          # If current subarray        # length satisfies        if(total_cost <= c):            return True          # If it is not possible    return False  # Driver CodeA = [1, 2, 1, 6, 5, 5, 6, 1]B = [14, 8, 15, 15, 9, 10, 7, 12]N = len(A)C = 40print(maxLength(A, B, N, C))  # This code is contributed by avanitrachhadiya2155

## C#

 // C# program for the above approachusing System;using System.Collections.Generic;  public class GFG {      // Function to find maximum length    // of subarray such that sum of    // maximum element in subarray in brr[] and    // sum of subarray in []arr * K is at most C    public static int maxLength(        int []a, int []b, int n, int c)    {        // Base Case        if (n == 0)            return 0;          // Let maximum length be 0        int max_length = 0;          int low = 0, high = n;          // Perform Binary search        while (low <= high) {              // Find mid value            int mid = low + (high - low) / 2;              // Check if the current mid            // satisfy the given condition            if (possible(a, b, n, c, mid)) {                  // If yes, then store length                max_length = mid;                low = mid + 1;            }              // Otherwise            else                high = mid - 1;        }          // Return maximum length stored        return max_length;    }      // Function to check if it is possible    // to have such a subarray of length K    public static bool possible(        int []a, int []b, int n, int c, int k)    {          // Finds the maximum element        // in each window of size k        List<int> dq            = new List<int>();          // Check for window of size K        int sum = 0;          // For all possible subarrays of        // length k        for (int i = 0; i < k; i++) {              sum += a[i];              // Until deque is empty            while (dq.Count > 0                   && b[i] > b[dq[dq.Count - 1]])                dq.RemoveAt(dq.Count - 1);            dq.Add(i);        }          // Calculate the total cost and        // check if less than equal to c        int total_cost = sum * k                         + b[dq];        if (total_cost <= c)            return true;          // Find sum of current subarray        // and the total cost        for (int i = k; i < n; i++) {              // Include the new element            // of current subarray            sum += a[i];              // Discard the element            // of last subarray            sum -= a[i - k];              // Remove all the elements            // in the old window            while (dq.Count > 0                   && dq                          <= i - k)                dq.RemoveAt(0);              while (dq.Count > 0                   && b[i]                          > b[dq[dq.Count - 1]])                dq.RemoveAt(dq.Count - 1);              dq.Add(i);              // Calculate total cost            // and check <=c            total_cost = sum * k                         + b[dq];              // If current subarray            // length satisfies            if (total_cost <= c)                return true;        }          // If it is not possible        return false;    }      // Driver Code    public static void Main(String[] args)    {        int []A = { 1, 2, 1, 6, 5, 5, 6, 1 };        int []B = { 14, 8, 15, 15, 9, 10, 7, 12 };          int N = A.Length;          int C = 40;          Console.WriteLine(            maxLength(A, B, N, C));    }}  // This code is contributed by Amit Katiyar

## Javascript

 

Output:

3

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Related Topic: Subarrays, Subsequences, and Subsets in Array

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